Consider an anharmonic oscillator in quantum mechanics, described by the Hamiltonian $$H=\frac{p^2}{2m}+\frac{1}{2}m\omega^2x^2+bx^4.$$ The $bx^4$ term doesn't cause scattering. The effect of this term only causes the oscillator eigenstates and energy eigenvalues to be modified.

On the other hand, a theory specified by the Lagrangian density $$\mathscr{L}=\frac{1}{2}(\partial_\mu\phi)^2-\frac{1}{2}m^2\phi^2-\frac{\lambda}{4!}\phi^4$$ with $m^2>0$ and $\lambda>0$ is interpreted as a self-interacting field of mass $m$. Here the term $\lambda \phi^4$ causes scattering! But there must again be noninteracting states which should at least be solvable perturbatively and which do not scatter off each other?

Does the first case i.e., $\phi^4$ theory fall within the purview of time-dependent perturbation theory and the latter within time-independent perturbation theory?

  • $\begingroup$ The theories are totally analogous. In the QM case you have a free massive boson field $x(t)$ perturbed by a quartic potential $V(x) = -b x^4$. In the QFT case you have a free massive boson perturbed by a quartic interaction $V(\phi) = - \frac{\lambda}{4!}\phi^4. In either case you have a free theory and you can do perturbation theory about the free field vacuum. $\endgroup$ – d_b Dec 15 '18 at 12:33

"Scattering" implies transition of probability amplitude from one eigenstate of the $\phi$ field to another. In the same sense, for perturbative treatment (just as in $\phi^4$) $x^4$ "scatters" the eigenstate of energy $E$ (of the free part) to energy $E'$ (state which was not accessible for same initial conditions with evolution solely due to the noninteracting part). $\phi^4$ as $x^4$ perturbation couples states, otherwise decoupled from the free part of Hamiltonian / Lagrangian (In analogy with Rabi oscillations, however here without external field to cause the transitions)

If, in principle, you could solve the $\phi^4$ model and set your initial condition to any of its eigenstates - you won't scatter anywhere.

  • $\begingroup$ We cannot solve the $\phi^4$ theory exactly but that doesn't mean we the exact eigenstates of $\phi^4$ theory do not exist. We could have taken the attitude that the eigenstates of $\phi^4$ theory are the particles and then there will be no scattering! What dictates that I have to declare the quanta of $\mathscr{L}_0=\mathscr{L}-\frac{\lambda}{4!}\phi^4$ as the particles of nature? @Alexander $\endgroup$ – SRS Dec 16 '18 at 14:33

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